Mathematics - 2017-08-31 (page 3 of 3)

what is wrong with this latex code

let (G,*) be a set with an associative binary operation.
\\1) $\exists$ an element e $\in$ G such that xe=x for all x $\in$ G
2)$ for every x $\in$ G $\exists$ a y $\in$G such that xy=e$

the first line 1) shows up fine

@KasmirKhaan Add a space after commands

but after 2) its red

give me an example of command pls

oh no

There is no such thing as \suchthat

hmm

5:07 PM

@KasmirKhaan it's \forall

let me send you an SS where the first line shows up fine

=p

@LeakyNun Umm, is that the same?

Faust

use | or s.t

@Mr.Xcoder no, I just saw $forevery$

I use :

Oh, that is the same :)

I use \text{such that}

5:08 PM

you see that the first line was fine

because you inverted everything

@KasmirKhaan That's Tex, not mathjax

Oh no

you put "for every x" in math mode and \in in normal mode

Yes ^

><

Dont get it

what did I invert?

let me delete the first

$

5:10 PM

math mode and normal mode

yes, exactly

You have put $...math...$ such that $...math...$ instead of $...math... such that ...math...$

okay let me try it that way =p

sorry guys its my first day on latex

2) for every x $\in$ G $\exists$ a y $\in$ G such that $xy=e$

still did not show up

show us what you typed

Did you download amsmath

5:14 PM

@PVAL-inactive no

@PVAL-inactive miktex

I think ) might be a protected character

just use 2.

okay let me try that :D

it worked on the first line

1)

still gives error =p

! Missing \endcsname inserted.

\end{documenxt\in}

it is there

it should be \end{document}

5:18 PM

@LeakyNun Did you learn tex (I have seen you on PPCG using it) for your maths studies to follow?

@Mr.Xcoder you can say so

\begin{enumerate}

look at the errors down =p

\item "stuff"

@KasmirKhaan you still haven't fixed it

5:20 PM

@KasmirKhaan Use \end{document} instead of \end{documenxt\in}

\end{enumerate}

\end{documenxt\in} makes no sense at all

okay

YEEEEEEES

:D

I have no idea how that got in there

BTW what ancient version of windows are you using?

"in" part

windows7

but i changed the view for better performance

like all the options for better view deleted =p

5:22 PM

Ok, I have to go!

o/

Okay =p thanks for help !

@KasmirKhaan \begin{enumerate} \item testing \end{enumerate}

@LeakyNun where should I put that?

where you used 1) and 2)

it is ficed now btw

5:23 PM

I know

ah okay

Don't use symbols when words explain things more clearly.

$\exists$ and $\forall$ have no business in mathematical writing unless they are a part of the objects being studied.

Some random mathmode tips: never use double dollar signs. Use [ ... ] or \ begin{equation*} ... \ end{equation*}

Errr

Guys, he's new to all of this.

one thing at a time :P

haha am just happy that those symbols just came out =P

But ill change it once its all done

:)

5:26 PM

Gotta learn good habits early

how do I change that second 1 to a 2 ?

you don't use begin enumerate every time :P

\begin{enumerate} \item lorem \item ipsum \end{enumerate}

\begin{enumerate}
    \item lorem
    \item ipsum
\end{enumerate}

aaaaaa ><

what is lorem and ipsum

words

nothing more

ah ><

5:30 PM

Also, technically it's preferrable to use \ ( ... \ ) for inline math instead of dollar signs, but they'll render the same so it's not that important.

@SteamyRoot what's the difference between double dollar and \[?

@SteamyRoot its my first day on latex ;D

@KasmirKhaan I know. But it's good to learn these things soon. Before you make a habit out of some bad practises.

@LeakyNun dollar signs are TeX, backslash-brackets are LaTeX

Yes yes :D

@SteamyRoot I see

5:32 PM

There's a lot of things wrong with double dollar signs, like vertical spacing

LaTeX notation will usually provide better (more specific) error messages if you do something wrong, and it's future-proof in case significant updates to LaTeX happen

@SteamyRoot I see

@KasmirKhaan that's way better

but I would put math symbols in math mode :P

:D

there exists an element $e \in G$ such that $xe = x$ for all $x \in G$

5:35 PM

ah okay =p

let me fix that

that looks way complicated than the dollar sign

it's definitely more annoying to type, yes

Fawad

Hi

Why this is called circled intersect internally?

because one circle is in the internal of the other...

Fawad

Hmm but ratio divides it externally 😐

Confusion lol

oups noticed soemthing missing

=P

5:45 PM

hmm?

i changed

new pic in a sec

$TeX - LaTeX$ TeX, LaTeX and Friends

\begin{chat}

9

6

maybe you would want to go here?

oh thanks :D

ill def go there =p

its looks way more pretty with ( stuff )

oups

meant this \ ( ....)

6:04 PM

@KasmirKhaan how is it

@LeakyNun Alot better now thanks :D

@KasmirKhaan when will you finish it?

@LeakyNun I can finish 2 or 3 today, the 4th did not solve yet =p

@KasmirKhaan I mean, how many hours later?

@LeakyNun Maybe 2 hours :D maximum 3

6:06 PM

hmm, then I should go to sleep now

it's 2:06 AM now

oh

omg =p yeah you should sleep =P

ok, see you

Goodnight and thanks alot for all your help! ill email the solution @LeakyNun

I mean the papper so you can correct me :D

@KasmirKhaan I'll check it as soon as I wake up

Okay :)

Blue Apple

6:15 PM

hi every1, if S $\superset$ R , then is $R^n$ $\otimes$ $S$ = $S^n$ true, if I take special example of Z and Q then it verifyes but in general what can we say?

@BlueApple equal as what? as $R$-modules?

Blue Apple

yes R modules

Then yes, if $a_1,\dots,a_n$ is a basis for $R^n$ then $a_i\otimes 1$ should be a basis for the tensor product

6:50 PM

Hi, I have a question about defining random variable. Suppose I said: define a random variable X=: 1 if algorithm outputs wrong answer, and 0 otherwise. and When we say Pr[X=k] = p (1-p)^k, then this is wrong?!

@YOUSEFY How could it be correct? The variable can never take values apart from $0$ or $1$

so you mean that X=1 or X=0!

yes, those are the two possible values by your definition

does the definition that I wrote means: Pr[X=1]=p, Pr[X=0] = 1-p, Pr[X=2]=0, Pr[X=3]=0, etc. Is this correct?

yes, where $p$ is the probability that the algorithm outputs the wrong answer

6:56 PM

I'm trying to define Bernolli random variable in order to use Geometric Distribution!

Suppose I define an event: E:= is the event where we have number of iterations of algorithm. Then, Pr[E=k] = p (1-p)^k (where k means the number of iterations). Is this valid definition!

So we will stop when we have a correct answer "probability p".

7:18 PM

@TobiasKildetoft Thank you Tobias. I understand now. It really helped me a lot. Forget about what I said in last comment. I think the probability Pr[E=k] should explain what algorithm do. Usually in las vegas algorithm we make the algorithm run forever and stops only when it has an answer. So, this is why we have a geometric distribution. I had a difficult time to define the correct random variable or event for this distribution. So, I came here to see how my definition works.

Thank you!

Lozansky

7:45 PM

Is there a theorem that says that the flux of a vector field over a region containing a singular point is equal to the flux of the vector field over a sphere centered at the singular point?

Given that the divergence of the vector field is $0$ in said region

I guess it's just the divergence theorem

Waiting

8:21 PM

@Hippalectryon o/ How is it going lately?

\o @Waiting

@skullpatrol Hey!

How are you? pal :-)

@skullpatrol In general working hard on my projects. How about you? :-)

skullpatrol

Fine, thanks.

Gotta run, cya @Waiting

Waiting

8:33 PM

@skullpatrol Okay! Take care.

skullpatrol

You too :-)

Balarka Sen

@Lozansky Yeah, it's Stokes' theorem over the region sandwiched between the boundary of the region and the sphere.

user193319

8:46 PM

Is is true that $m^*(A-B) = m^*(A) - m^*(B)$, where $m^*(A) = \inf \{ \sum_{k=1}^\infty \ell (I_k) ~|~ A \subseteq \bigcup I_{k=1}^\infty I_k \}$ denotes the outer measure on $\Bbb{R}$?

Hippalectryon

@Waiting o/ fine, and you ?

Waiting

@Hippalectryon That's good. Not that bad here, working on my stuff as usual. I thought to enter a bit the room and see what the news is.

Lozansky

9:29 PM

If a curve $L$ is an equipotential curve to a scalar field $\phi$, then surely $\nabla \phi = \mathbf{0}$ for points on that curve?

@BalarkaSen Observation

People tend to get annoyed when we start to go off on tangents unrelated to math. But tangents are part of math, aren't they?

Like in geometry, or trigonometry :P

Balarka Sen

For a minute I thought you were going to tell me a mathematical observation

I guess you got me annoyed by going off on a tangent

This is meta on multiple levels

9:42 PM

@Lozansky definitely not

Take $\phi=x^2+y^2$. The equipotential curves in the plane are then circles, and the gradient only is zero at the origin

There is a relation between the gradient at a given point and the equipotential through it, but it's not "gradient is zero"

Lozansky

10:10 PM

@Semiclassical Oh right. But $\nabla \phi \cdot d\mathbf{r} = 0$, no?

Mary Star

We have the double integral $$ I=\int_0^1 \int_2^{2e^x}f(x,y) dydx $$
and I want to change the order of the integrals. I have done the following:

From the integral we have that $0\leq x\leq 1$ and $2\leq y\leq 2e^x$.

So, we get that $y\leq 2e^x \Rightarrow \frac{y}{2}\leq e^x \Rightarrow \ln \left (\frac{y}{2}\right )\leq x$.

Therefore, we get that $\ln \left (\frac{y}{2}\right )\leq x\leq 1$ and $2\leq y\leq 2e^x\leq 2e^1=2e$.

So, by changing the order of integrals we get the following $$I=\int_2^{2e} \int_{\ln \left (\frac{y}{2}\right )}^1f(x,y) dxdy $$

usukidoll

hallllllllo

I think I'm screwing up my order
let's say I have $ B \times D$ as a product topology
By product topology def there's a basis B collection of the form U x V
where B is an open subset of U and D is an open subset of V. Do I type it like
$B \times D \subset U \times V$
or
$U \times V \subset B \times D$

SORRY SLEEP DEPRIVED :O

I think it's the second one..first one is like U is an open subset of B um wut

10:30 PM

@Lozansky if r(t) parametrizes the equipotential, yeah

usukidoll

yeah it's the second one. just reread

10:52 PM

4 hours ago, by YOUSEFY

so you mean that X=1 or X=0!

more like, "X=0 or X=1!"

2 hours ago, by user193319

Is is true that $m^*(A-B) = m^*(A) - m^*(B)$, where $m^*(A) = \inf \{ \sum_{k=1}^\infty \ell (I_k) ~|~ A \subseteq \bigcup I_{k=1}^\infty I_k \}$ denotes the outer measure on $\Bbb{R}$?

@user193319 Take $A=[0,2]$ and $B=[3,4]$. Obviously, the LHS equals 2 and the RHS equals 1.

@MaryStar That's correct, but remember to check the condition $0 \le x$ next time. (It's a coincidence that you don't need to check it this time, because $\ln \left( \frac y 2\right) \ge 0$.)

Mary Star

@LeakyNun Ah ok!! Thank you!! :-)

@MaryStar no problem

11:09 PM

@LeakyNun Did you sleep ? =P

@LeakyNun I sent you btw the pdf :D

@KasmirKhaan for 4.5 hours

:D

you cant sleep more?

should I send you the remarks by email or here?

you should get at least 6

remarks here are fine

but the papper dotn send it anywhere

sure

11:11 PM

because maybe my teacher thinks i copied from somewhere ><

well, you used $x^{-1}$ without defining it

okay ill copy the remakrs in sepeart file

so i keep track of them =p

and use this:

\[\begin{array}{rcll}
1+1 & = & 2 & \text{Peano} \\
2+2 & = & 4 & \text{double both sides}
\end{array}\]

and obvious typo on the first line...

and you can state the third line on its own... it's like the definition of $x^{-1}$

@AkivaWeinberger hi

Hi

let me try it

11:14 PM

If I have a vector is there a notation that turns it into a diagonal matrix?

Like say $v=[v_1,\dots,v_n]$, and $V=\begin{bmatrix}v_1&&\\&\ddots&\\&&v_n\end{bmatrix}$

ahh more fancy way to put the comments nice

:D

how do you relate $v$ and $V$

I saw diag(v1,...,vn) on wikipedia

Hm, so ${\rm diag}(\mathbf v)$ could also make sense

@AlessandroCodenotti hi

@AkivaWeinberger $\operatorname{diag}(\vec v)$!

habits :P

\operatorname

Alessandro Codenotti

11:18 PM

diag is also a matlab commanf that does exactly that iirc

Hm and multiplying it by $[1,1,\dots,1]^\top$ undoes it

@LeakyNun am gonna sleep its 2 am here almost

Do.

@LeakyNun can we continue in the morning ? :)

how do you say "undoes"?

un+does? undo+s?

Alessandro Codenotti

11:19 PM

(As well as the opposite, extracting the diagonal of a matrix as a vector, if given a matrix as input)

@LeakyNun Un+does

@KasmirKhaan, your approach is completely wrong

Unduzz, not undooz :P

@LeakyNun damn it >< what was wrong ?

Arright sweet dreams @KasmirKhaan

11:20 PM

you skipped steps.

@LeakyNun can you please send me the comment on email ? like you paint and put the wrong on a red box =p

really sleepy now

@KasmirKhaan can I just put the whole thing on a red box

been working on that thing for 4 hours

><

:((((((((

damn it

send me on email

ill look at it first ghing

I don't know what you're trying to do but sleep is a good idea

when i wake up =p

haha

thanks alot @LeakyNun and good night yall :D

11:22 PM

@KasmirKhaan good night

11:39 PM

@BalarkaSen why is sphere eversion possible?

@LeakyNun here?

@KasmirKhaan there, everywhere

@LeakyNun got the ping from the email on my phone

><

Ehm

I could not read what you wrote

I mean nothing is there

??

what do you mean by nothing is there

I wrote like using a red pen everywhere

oups ><

yeah I see now =p

hmm

the second comment

if that is not associativity then what is it?

11:47 PM

you didn't use it at all

you changed nothing

and you don't know it because you didn't use brackets

when you multiply $xx^{-1}$ on the right by $x$, you get $(xx^{-1})x$.

you need the brackets.

this is werid

they dont show for you but i did use them

let me show u an SS of texstudio

if you used the brackets then you would have known that you didn't apply associativity @_@

I don't see the brackets there

5th line of the proof

they are ther

11:51 PM

...

I'm talking about the 4th line

2 mins ago, by Leaky Nun

when you multiply $xx^{-1}$ on the right by $x$, you get $(xx^{-1})x$.

2 mins ago, by Leaky Nun

you need the brackets.

yes thats what i meant

@Semiclassical hi

hi

I used associativity to do xx' first

@KasmirKhaan ...

11:52 PM

(xx')x

i did it this way

that isn't associativity

hmm

multiplying xx' on the right by x gives you exactly (xx')x.

what is it then

no associativity needed.

11:52 PM

I know

but doing xx' first

should I just

what... that's just standard multiplication

ahhh

okay now I see what you mean

grrrrrr

How can I make the parentses look better

they look very thin

@KasmirKhaan are they bad?

its hard to see them

\mathbf{(} yx \mathbf{)}

11:55 PM

let me try that :D

where do you keep getting those commands ;D

experience

@Semiclassical d'you wanna see a magic trick?

anon

"help displaying a formula" on wikipedia is a nice reference

@LeakyNun the part when i said yxy =y , and i said because i prove ex=x for all x

why is that not relevant?

yxy=y => yx=e

this step, you need justification

I don't see how "ex=x" is justification

hmm what i meant by that is that we have ex=x for all x

means that the element yx dotn change the other elemetns when multiplied on the left

so it is equal to e

11:59 PM

Kasmir

Yes?